Problem: Graph this system of equations and solve. $2x+y = -1$ $12x+2y = 6$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Convert the first equation, $2x+y = -1$ , to slope-intercept form. $y = -2 x - 1$ The y-intercept for the first equation is $-1$ , so the first line must pass through the point $(0, -1)$ The slope for the first equation is $-2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) $1$ position to the right. $2$ positions down from $(0, -1)$ is $(1, -3)$ Graph the blue line so it passes through $(0, -1)$ and $(1, -3)$ Convert the second equation, $12x+2y = 6$ , to slope-intercept form. $y = -6 x + 3$ The y-intercept for the second equation is $3$ , so the second line must pass through the point $(0, 3)$ The slope for the second equation is $-6$ . Remember that the slope tells you rise over run. So in this case for every $6$ positions you move down (because it's negative) $1$ position to the right. $6$ positions down from $(0, 3)$ is $(1, -3)$ Graph the green line so it passes through $(0, 3)$ and $(1, -3)$ The solution is the point where the two lines intersect. The lines intersect at $(1, -3)$.